Chain rule of partial derivatives!

Homework Statement

Suppose f(x,y) = 2x^5 + 4xy + 2y^3
g1(u,v) = u^2 - v^2
g2(u,v) = uv
h(u,v) = f(g1(u,v), g2(u,v))

Use chain rule to calculate:
dh/du (1,-1) and dh/dv (1,-1)

The Attempt at a Solution

i let h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) + 2(u-v)^3
then i tried drawing a chain rule tree diagram:

h --> f --> g1 and g2

g1 --> x and y

g2 --> x and y

so dh/du = (dh/df . df/dx . dx/du) + (dh/df . df/dy . dy/du)

then i got stuck :S

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vela
Staff Emeritus
Homework Helper
Stuck how?

im not sure if my chain rule is correct to begin..i found this in my textbook though on calculating partial derivatives with changes in variables:

dh/du = dh/dg1 . dg1/du + dh/dg2 . dg2/du

and got this

since h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) + 2(u-v)^3

dh/dg1 = 10(u^2-v^2)^4 + 4uv

dg1/du = v

is that right so far?

vela
Staff Emeritus
$$h = f(g_1, g_2) = 2g_1^5 + 4g_1g_2 +2g_2^3$$so$$\frac{\partial h}{\partial g_1} = 10g_1^4 + 4g_2$$and then express the g's in terms of u and v. (This gives you what you got above.) Do the same thing with g2. Then put those together with the partial derivatives of the g's to form the final result. (You got $\partial g_1/\partial u$ incorrect.)